What is the chain rule in simple terms?

The chain rule says that to differentiate a composite function — a function inside another function — you multiply the derivative of the outer function (evaluated at the inner function) by the derivative of the inner function. In words: outer rate × inner rate.

When do I use the chain rule?

Use it whenever you have a "function of a function", such as sin(x²), e^(3x), or (2x+1)⁵. If you can spot an inner expression that has been substituted into an outer function, the chain rule applies.

What is the derivative of sin(x²)?

By the chain rule, the outer function sin has derivative cos, and the inner function x² has derivative 2x. So d/dx sin(x²) = cos(x²) × 2x = 2x·cos(x²).

Why do you multiply the two derivatives?

Because a small change in x first changes the inner function at rate g′(x), and that change then changes the outer function at rate f′(g(x)). The two rates compound, so the overall rate is their product — exactly like gear ratios multiplying.

What is the most common chain rule mistake?

Forgetting to multiply by the derivative of the inner function. Writing d/dx sin(x²) = cos(x²) misses the ×2x factor. Always differentiate the inside as well.

Chain Rule Explained (Interactive) — Differentiate Composite Functions

The chain rule is how you differentiate a function of a function — expressions like $\sin(x^2)$ , $e^{3x}$ or $(2x+1)^5$ . Most students learn it as a formula to memorise. Here we’ll build the intuition first: once you see whytwo rates multiply, you’ll never forget the “ $\times g'(x)$ ” again.

The core intuition: rates multiply

Imagine two connected gears. Turn the first gear and it drives the second. If the first turns twice as fast as your hand, and the second turns three times as fast as the first, then the second turns 2 × 3 = 6 times as fast as your hand. The rates compound — they multiply.

A composite function $h(x) = f(g(x))$ is exactly this chain of gears. Nudge $x$ a tiny bit. First, the inner function $g$ responds at its own rate, $g'(x)$ . That change in $g$ then drives the outer function $f$ , which responds at its rate, $f'(g(x))$ . The overall rate of change is the product of the two.

The rate of a composite is the outer rate times the inner rate.

See it: the two rates multiplying

Drag the point along the curve below. The panel breaks the slope of the tangent into its two factors — the inner rate $g'(x)$ and the outer rate $f'(u)$ — and shows their product. Notice that the product always equals the actual slope of the tangent line. Try each example function.

InteractiveChain rule: rate decomposition

Loading interactive…

Drag the point (or use arrow keys). The product of the inner and outer rates equals the tangent's slope for every x.

Text description ↓

An interactive graph of a composite function $h(x)=f(g(x))$ with a draggable point and its tangent line. A readout decomposes the tangent’s slope into the inner rate $g'(x)$ and outer rate $f'(g(x))$ , whose product equals the slope. Available functions and their derivatives:

Function h(x)	Inner g(x)	Derivative h′(x)
sin(x²)	x²	2x·cos(x²)
e^(3x)	3x	3·e^(3x)
(2x+1)⁵	2x+1	10·(2x+1)⁴
√(x²+1)	x²+1	x / √(x²+1)

The rule, formally

In Lagrange (prime) notation, for $h(x) = f(g(x))$ :

\frac{d}{dx}\,f\big(g(x)\big) = f'\big(g(x)\big)\cdot g'(x)

Leibniz notation makes the “compounding rates” idea even clearer. Write $u = g(x)$ so that $y = f(u)$ . Then

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.

It looks like the $du$ ’s cancel like fractions. That’s a useful memory hook, but not a proof — $\frac{dy}{du}$ is a limit, not a fraction. The genuine justification is that the small changes satisfy $\Delta y \approx \frac{dy}{du}\,\Delta u$ and $\Delta u \approx \frac{du}{dx}\,\Delta x$ , and substituting one into the other gives the product in the limit.

The three-step recipe

Spot the layers. Identify the inner function $g(x)$ (what’s “inside”) and the outer function $f$ .
Differentiate the outside, keeping the inside untouched: $f'(g(x))$ .
Multiply by the derivative of the inside, $g'(x)$ .

Worked examples

1Worked example — a trig composite

Differentiate $h(x) = \sin(x^2)$ .

Inner: $g(x) = x^2$ , so $g'(x) = 2x$ . Outer: $f(u) = \sin u$ , so $f'(u) = \cos u$ . Assemble:

h'(x) = \cos(x^2)\cdot 2x = 2x\cos(x^2).

2Worked example — an exponential

Differentiate $h(x) = e^{3x}$ .

Inner $g(x)=3x$ gives $g'(x)=3$ ; outer $f(u)=e^u$ gives $f'(u)=e^u$ . Hence

h'(x) = e^{3x}\cdot 3 = 3e^{3x}.

3Worked example — a power (the general power rule)

Differentiate $h(x) = (2x+1)^5$ . Inner $g(x)=2x+1$ , $g'(x)=2$ ; outer $f(u)=u^5$ , $f'(u)=5u^4$ . So

h'(x) = 5(2x+1)^4\cdot 2 = 10(2x+1)^4.

This is the “general power rule”: $\frac{d}{dx}\,[g(x)]^n = n[g(x)]^{n-1}g'(x)$ — just the chain rule with an outer power.

Common mistakes

Practice

Your turn

Differentiate $h(x) = \cos(5x)$ .

Show the answer ↓

Inner $5x$ has derivative 5; outer $\cos$ differentiates to $-\sin$ . So $h'(x) = -5\sin(5x)$ .

Your turn

Differentiate $h(x) = (3x^2 - 1)^4$ .

Show the answer ↓

Outer power: $4(3x^2-1)^3$ . Inner derivative: $6x$ . Product: $h'(x) = 24x(3x^2-1)^3$ .

The core intuition: rates multiply

The rate of a composite is the outer rate times the inner rate.

See it: the two rates multiplying

InteractiveChain rule: rate decomposition

Loading interactive…

Drag the point (or use arrow keys). The product of the inner and outer rates equals the tangent's slope for every x.

Text description ↓

Function h(x)	Inner g(x)	Derivative h′(x)
sin(x²)	x²	2x·cos(x²)
e^(3x)	3x	3·e^(3x)
(2x+1)⁵	2x+1	10·(2x+1)⁴
√(x²+1)	x²+1	x / √(x²+1)

The rule, formally

In Lagrange (prime) notation, for $h(x) = f(g(x))$ :

\frac{d}{dx}\,f\big(g(x)\big) = f'\big(g(x)\big)\cdot g'(x)

Leibniz notation makes the “compounding rates” idea even clearer. Write $u = g(x)$ so that $y = f(u)$ . Then

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.

The three-step recipe

Spot the layers. Identify the inner function $g(x)$ (what’s “inside”) and the outer function $f$ .
Differentiate the outside, keeping the inside untouched: $f'(g(x))$ .
Multiply by the derivative of the inside, $g'(x)$ .

Worked examples

1Worked example — a trig composite

Differentiate $h(x) = \sin(x^2)$ .

Inner: $g(x) = x^2$ , so $g'(x) = 2x$ . Outer: $f(u) = \sin u$ , so $f'(u) = \cos u$ . Assemble:

h'(x) = \cos(x^2)\cdot 2x = 2x\cos(x^2).

2Worked example — an exponential

Differentiate $h(x) = e^{3x}$ .

Inner $g(x)=3x$ gives $g'(x)=3$ ; outer $f(u)=e^u$ gives $f'(u)=e^u$ . Hence

h'(x) = e^{3x}\cdot 3 = 3e^{3x}.

3Worked example — a power (the general power rule)

Differentiate $h(x) = (2x+1)^5$ . Inner $g(x)=2x+1$ , $g'(x)=2$ ; outer $f(u)=u^5$ , $f'(u)=5u^4$ . So

h'(x) = 5(2x+1)^4\cdot 2 = 10(2x+1)^4.

This is the “general power rule”: $\frac{d}{dx}\,[g(x)]^n = n[g(x)]^{n-1}g'(x)$ — just the chain rule with an outer power.

Common mistakes

Practice

Your turn

Differentiate $h(x) = \cos(5x)$ .

Show the answer ↓

Inner $5x$ has derivative 5; outer $\cos$ differentiates to $-\sin$ . So $h'(x) = -5\sin(5x)$ .

Your turn

Differentiate $h(x) = (3x^2 - 1)^4$ .

Show the answer ↓

Outer power: $4(3x^2-1)^3$ . Inner derivative: $6x$ . Product: $h'(x) = 24x(3x^2-1)^3$ .

The Chain Rule, Made Intuitive

The core intuition: rates multiply

See it: the two rates multiplying

The rule, formally

The three-step recipe

Worked examples

Common mistakes

Practice

Frequently asked questions

Keep exploring

Want a tutor to walk you through it?

The Chain Rule, Made Intuitive

The core intuition: rates multiply

See it: the two rates multiplying

The rule, formally

The three-step recipe

Worked examples

Common mistakes

Practice

Frequently asked questions

Keep exploring

Want a tutor to walk you through it?